Chap1 slides (2)

August 2012 1-2Normal Stress ten-sile stress.. psi Pounds per square inch lb/in.2 ksi Kilopounds kips per square inch 103 lb/in.2 kPa Kilopascal 103 N/m2 MPa Megapascal 106 N/m2 GPa Giga

Stress

member

Learning objectives

forces and moments

Static

August 2012 1-2

Normal Stress

ten-sile stress

com-pressive stress

psi Pounds per square inch lb/in.2

ksi Kilopounds (kips) per square inch 103 lb/in.2

kPa Kilopascal 103 N/m2

MPa Megapascal 106 N/m2

GPa Gigapascal 109 N/m2

Tensile Normal Force

Compressive Normal Force

Imaginary Cut

Chandelier Weight

Building Weight

Imaginary Cut

Chandelier Weight Tensile Normal Stress

Building Weight

Compressive Normal Stress

N

σavg

σavg σavg σavg

C1.1 A 6 kg light shown in Fig C1.1 is hanging from the ceiling by wires of diameter of 0.75 mm Determine the tensile stress in the wires

Fig C1.1

Light

2.5 m

2.5 m

2 m

B C

August 2012 1-4

Shear Stress

Weight

of the Clothes Imaginary cut

between the wall

and the tape

V

Weight

of the Clothes

τ τ

Imaginary cut along the possible path

of the edge of the ring.

Pull

of the hand

Pull

of the hand

Pull

of the hand

τ

τ

τ

V V V

Mwall

Clothes

Shear stress in pins

V

F

V V

F

F

F

B

D

N B

N D

B

C

D

Cut 1

N B

N C

N D

V B

V D

Cut 2

C

N C

V B

V D

Multiple forces on a pin

August 2012 1-6

C1.2 The device shown in Fig C1.2 is used for determining the shear strength of the wood The dimensions of the wood block are

6 in x 8 in x 1.5 in If the force required to break the wood block is

12 kips, determine the average shear strength of the wood

Fig C1.2

P

Wood

6 in

6 in 2 in

C1.3 Two cast iron pipes are held together by a bolt as shown The outer diameters of the two pipes are 50 mm and 70 mm and wall thick-ness of each pipe is 10 mm.The diameter of the bolt is 15 mm The bolt broke while transmitting a torque of 2 kN-m On what surface(s) did the bolt break? What was the average shear stress in the bolt on the surface where it broke?

Fig C1.3

August 2012 1-8

C1.4 Fig C1.4 shows a truss and the sequence of assembly of members at pin H All members of the truss have a cross-sectional area of

stresses in members HA, HB, HG and HC of the truss shown in Fig C1.4 (b)Determine the maximum shear stress in pin H

Fig C1.4

3 m

F

G H

HAHB

HG HC

Pin H

Internally Distributed Force System

of a body is called the stress on a surface

of an imaginary cut is called the normal stress on a surface

of an imaginary cut surface is called the shear stress on the surface.

A

B C D E

(a)

(a)

F B

F C

F E

F D

F A

␴ A

␶ A

B E

D C

A

Tangent in plane

(b) Normal to plane

Static

x

y z

x

y z

x

y z

x

y z

M z

T

M y

Normal stress linear in y

Normal stress linear in z

Uniform shear stre

in tangen direction Uniform Normal

Stress σavg

N = σavgA V = τavgA

Uniform Shear Stress τavg

August 2012 1-10

C1.5 An adhesively bonded joint in wood is fabricated as shown in Fig C1.5 The joint is to support a force P = 25 kips, what should be the length L of the bonded region if the adhesive strength in shear is 300 psi

Fig C1.5

L

8 in

1 in

P

P

Class Problem 1

In problems below, draw the free body diagram (FBD) that can be used for calculation of shear stress Identify the surface and the direction of shear stress

2 in.

P

12 in.

hammer Assuming the nail does not bend

or break and the hammer does not slip

Show the shear stress on the nail in the

FBD

bonded Show the shear stress in the

adhesive in the FBD

togather Show the shear stress in the

bolts in the FBD

August 2012 1-12

Stress at a Point

in the positive i direction

Table 1.1 Comparison of number of components

i

direction of outward normal to the imaginary cut surface.

direction of the internal force component.

i

ΔAlimi → 0

=

Stress Element

point by constructing surfaces that have outward normal in the coordi-nate directions

Construction of a Stress Element for Axial Stress

Stress components are distributed forces on a surface .

x y

z

P P

y

A

σxx

B

x

z

σxx

y

σxx

x

z

σxx

August 2012 1-14

Construction of a Stress Element for Plane Stress:

All stress components on a plane are zero

Symmetric Shear Stresses:

from the corner

Stress cube showing all positive stress components

x

z

y

dz

dy

␴ xx

␴

␴ xx

␴

␶ xy

␶

␴ yy

␴␴

␴ yy

␴␴

␶ yx

␶␶

dx

A B

C

D

A B

C

y

σxx

σxx

σyy

σyy

τxy

τxy

τyx

τyx

dy

dx

3-dimensional element 2-dimensional element

C1.6 Show the non-zero stress components on the A,B, and C faces

of the cube

Class Problem 2

Show the non-zero stress components of problem C1.6 on the A,B, and C faces of the cube below

y

x

z A

.B C

z y

x

.A

.B C

August 2012 1-16

C1.7 Show the non-zero stress components in the r, θ, and x cylin-drical coordinate system on the A,B, and C faces of the stress element shown

x θ

r

A

B C